A 300 g wooden block on a smooth, level surface is firmly attached to a very light horizontal spring with a spring constant of 200 N/m.
The spring-block system is initially stretched a distance “d” beyond the spring's equilibrium position and is released from rest. As the block moves to the right, its speed is 50.0 cm/s when the spring-block system is 4.00 cm from the equilibrium position.
Find d, the initial displacement of the spring-block system (in cm).

The oscillatory periodic motion of a system occurs when there is a recovered force, in the special case that this force is proportional to the displacement is called simple harmonic motion, which is described by the expression.

x = A cos (wt + Ф)

w² = k/m

where x is the displacement, A the amplitude, w the angular velocity, t the time, k the spring constant, m the mass, and Ф a phase constant determined by the initial conditions.

Let's find the angular velocity/

w= [tex]\sqrt{ \frac{200}{0.300} }[/tex]

w = 25.8 rad/s

Let's look for the constant Ф, as the system is released from rest its initial velocity is zero, for zero time. The definition of speed is:

v= [tex]\frac{dx}{dt}[/tex]

v= - A w sin (wt +Ф)

0 = -A w sin Ф

Ф= 0

They indicate that at a given instant of the time the velocity is v= 50.0 cm/s and it is in a position x= 4.00 cm, let us write the equations for this time

Position.

4.00 = A cos 25.8t

Speed.

50.0 = - At 25.8 sin 25.8t

To solve the system, ;et's square and add.

Cos² 25.8t = [tex]\frac{16}{A^2}[/tex]

sin² 25.8t = [tex]\frac{3.756}{A^2 }[/tex]

1 = [tex]\frac{1}{A^2} \ (16 + 3.756)[/tex]

A = [tex]\sqrt{19.756}[/tex]

A= 4.44 cm

In conclusion using the solution for the oscillatory movement we can find the result for the amplitude of the initial displacement is:

The range of motion is: A = 444 cm

Learn more about oscillatory motion here: brainly.com/question/14311816